Eulerian description for a deforming control volume According to this document, One can find the following statement:

One concept is the control volume, which can be either finite or infinitesimal. Two types of control volumes can be employed:
1) Volume is fixed in space (Eulerian type).  Fluid can freely pass through the volume’s boundary.
2) Volume is attached to the fluid (Lagrangian type). Volume is freely
  carried along with the fluid, and no fluid passes through its
  boundary. This is essentially the same as the free-body concept
  employed in solid mechanics.

However, the control volume could be either: fixed, moving, or deforming.
I find the definition above confusing. Could we still describe the deforming control volume (e.g. deflating balloon) using Eulerian approach? if so, why it was defined as fixed?
I appreciate your help.
 A: "Could we still describe the deforming control volume (e.g. deflating balloon) using Eulerian approach?" The answer is very clearly No , based on the picture provided in the link you posted. As you can see, the diagram on the left that shows the Eulerian control volume clearly says fixed in space, so no question of deformation. To use your terminology, only the Lagrangian picture describes the motion of "deforming balloons".

The Eulerian picture describes fluid flow from the point of view of a fixed volume element in space. The Lagrangian picture describes fluid flow from the point of view of a single fluid element. I am not sure that I am using standard terminology here, but basically, what I mean is, say one drops a drop of dye in the fluid. The dye marks a single "fluid element". We can track its motion as the fluid flows. Importantly, in the context of this question, the fluid element is deformable. 
Also, it should not be hard to motivate that nothing flows in or out of the fluid element as the fluid evolves in time, as it is, by definition, the dyed section of the fluid at any given time. 
To make things clearer, one can try to derive the conservation laws using both pictures. I'll derive the conservation law for mass (which is the simplest).
First, the Eulerian Picture: Consider a fixed control volume $V$. The mass density is given by $\rho (x,t)$, hence, the total mass in $V$  at time $t$ is given by $M_V=\int_V dV \rho(x,t)$ . Mass is conserved in non-relativistic physics, hence it follows that the rate at which $M_V$ is increasing must equal the rate at which mass is flowing into $V$ ; mass can neither be created not be destroyed. Hence, $\partial_t (\int_V dV \rho) = - \int_{\partial V} \vec{dS}. \rho \vec{v}$. 
The RHS is a surface integral over the boundary of $V$ (denoted by $\partial V$) . The surface element is given by $\vec{dS}$ (direction is towards outward normal at the given point on the boundary). The fluid velocity is given by $\vec{v}$ . As such, we can see that the RHS is precisely the rate at which mass is flowing in. Using the divergence theorem, we can rewrite the RHS as a volume integral, finally giving us $\int_V [\partial_t \rho + \vec{\nabla}.(\rho \vec{v})] = 0$. If we now think of $V$ as an infinitesimal volume (small enough that $\rho,v$ can be, to a good approximation, taken to be constant across $V$), then we are led to the equation for conservation of mass, $ \partial_t \rho + \vec{\nabla}.(\rho \vec{v}) = 0$
Now, let us derive the same in the Lagrangian picture. In this picture, we think of a volume $V$ that is an infinitesimal fluid element (in the sense described above) of instantaneous volume $\Delta V$. Clearly, now, the rate at which the mass of the element is changing is $0$ (as nothing flows in or out of the Lagrangian fluid element), but the volume itself is changing.
We want to express the fact that mass of the fluid element does not change as the element flows with the fluid :  $\frac{d}{dt} M_V = \frac{d}{dt}(\int_V \rho) \approx \frac{d}{dt}(\rho \Delta V) =  0$. $\frac{d}{dt}$ denotes the change in any quantity as seen from the point of view of an observer sitting in the fluid element. 
We can see that $\frac{d}{dt} \rho = \partial_t \rho + \vec{v}.\vec{\nabla}{\rho}$ ; from the point of the observer in the fluid element, explicit variation of $\rho$ in time, and apparent variation in time (due to the fact that the observer is moving through regions of different $\rho$) both register on an equal footing. This is reflected in the two terms.
We also need to account for the variation in the volume. To find the rate of change in the volume, we need to compute the rate of deformation of the boundary of the volume. ie. (Abusing notation a bit) : $\frac{d}{dt} \Delta V= \frac{d}{dt} (\int_V) = (\int_{\partial V} \vec{dS}.\vec{v})  = \int_V \vec{\nabla}.\vec{v}  \approx \Delta V [\vec{\nabla}.\vec{v}]$, because each surface element is moving with the fluid velocity at that point. 
Now, we can put everything together and see that the equation $\frac{d}{dt}(\rho \Delta V) =  0$ reduces to $\Delta V \frac{d}{dt} \rho + \rho \frac{d}{dt} \Delta V = \Delta V [\partial_t \rho + \vec{v}.\vec{\nabla}{\rho} + \rho \vec{\nabla}.\vec{v}] = 0$ , which gives us the same equation for conservation of mass as above.
We can see that the Lagrangian picture is really an extension of mechanics of point particles, only now, the "particles" are extended objects that are deformable, and the forces on them are applied by a continuous medium (the surrounding fluid). A point particle obeys $\frac{d}{dt}m = 0$ , $F=ma$ , and $\frac{d}{dt} E = F.v$ , where $F$ is the (external) Force acting on $m$. These are exactly the same equations we can write down for the Lagrangian Fluid element.
The Eulerian picture is arguably more natural to think about fluids with. In particular, the equations obtained above are conceptually simpler to think of : Total mass, momentum and energy of the fluid are conserved. Correspondingly, we may write down conservation laws, which take the form $\partial_t (density) + \vec{\nabla}.(\vec{flux}) = 0$
